LeetCode题解:(114) Flatten Binary Tree to Linked List

题目说明

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

     1
    / \
   2   5
  / \   \
 3   4   6

The flattened tree should look like:

  1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6

题目分析

第一感觉是前序遍历,顺便打算在这题练习一下昨天学到的二级指针的写法XD,调的时候bug挺多的,可读性贼差,指针还是慎用啊……

以下为个人实现(C++,12ms):

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void flatten(TreeNode* root, TreeNode** &rightRef) {
        if (!root) return;
        *rightRef = root; // put root to right
        TreeNode* right = root->right; // store right ptr
        rightRef = &root->right; // move rightRef
        flatten(root->left, rightRef); // link left to right
        flatten(right, rightRef); // link right to the tail of left
        root->left = nullptr; // null left
    }
    
    void flatten(TreeNode* root) {
        TreeNode** right = &root;
        flatten(root, right);
    }
};

这是一种递归实现,讨论区看到了非递归实现,效率肯定要更高一些(C++,原帖):

void flatten(TreeNode *root) {
	while (root) {
		if (root->left && root->right) {
			TreeNode* t = root->left;
			while (t->right)
				t = t->right;
			t->right = root->right;
		}

        if(root->left)
		    root->right = root->left;
		root->left = NULL;
		root = root->right;
	}
}

这种算法思路是将结点的右分支(如果存在)放到左分支(如果存在)的最右边,然后再把左分支移动到右分支,因为任意一个结点至多被访问两次,所以时间复杂度是O(n)。

(LeetCode评测时间也不是这么靠谱的,试着提交了给出的8ms样例,结果还是12ms XD)

posted @ 2018-04-03 19:58  森高Slontia  阅读(129)  评论(0编辑  收藏  举报